Question

Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines: 

${\displaystyle \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\text{and}\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}}$

Answer 1

Let us assume the required line

${\displaystyle \overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})}$      .....(i)

The lines ${\displaystyle \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}}$ and ${\displaystyle \overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})}$ are perpendicular to each other.

The direction ratios of these lines are 3, −16, 7 and b₁, b₂, b₃. These lines are perpendicular to each other if

3b₁ − 16b₂ + 7b₃ = 0               .....(ii) 

Similarly, the direction ratios of the lines ${\displaystyle \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}}$ and ${\displaystyle \overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})}$ are 3, 8, −5 and b₁, b₂, b₃ are mutually perpendicular.

∴ 3b₁ + 8b₂ − 5b₃ = 0                   .....(iii)

From equations (ii) and (iii),

${\displaystyle \frac{b_1}{80-56}=\frac{b_2}{21+15}=\frac{b_3}{24+18}}$

and ${\displaystyle \frac{b_1}{24}=\frac{b_2}{36}=\frac{b_3}{72}}$

${\displaystyle \frac{{\text{b}}_1}{2}=\frac{{\text{b}}_2}{3}=\frac{{\text{b}}_3}{6}}$

Putting the proportional values ​​of b₁, b₂, b₃  (i) 

${\displaystyle \overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})}$

This is the equation of the required line.